 OK, so the point of last time was kind of crucial for our course. So remember, we have defined the Hessian of a function. And we studied, as a particular example, the Hessian of the height function, so surface, point, and the plane. OK, so remember, you have to make computations for critical points if you want to determine properties of the Hessian of a function. So in that case, critical points correspond to points where you are studying the height of a function based on a plane, which has a normal vector a, for example. And critical points we knew by geometric interpretations were exactly those points for which the normal vector of the surface at that point corresponds exactly to the normal vector to the plane that you are looking at. Thanks to this, we studied the Hessian. We identified the Hessian with the height function with respect to the plane with the second fundamental form. Very well. And then all the interpretation, Gauss curvature, and so on. The surface lies on one side of the tangent plane, crosses the tangent plane, and so on. So that was the moral of last lecture. Today, we play a similar game or with a different function. So we take our regular surface, s. We take a point outside or inside as usual. At the moment, it doesn't really matter anywhere in R3. And we look at the function f to be the distance squared. So f, actually let me write it here, f from s to r to be defined to be the function which, at a given point, gives you the norm squared of the vector p minus p0. So this is the distance squared. Now we know already what it means for a point to be a critical point for this function. So we gave the geometric interpretation also in this case. So remember, p is critical for f if and only if. Let's say it precisely. If and only if p0 lies in the normal line, so it's in the normal line at the point p for some lambda. So in particular, what I draw is not a critical point because the normal line here would go somewhere else. So here, if I want to guess where is a critical point, I should go more or less there. So how much is the action of this function at a point like this? Well, remember, so compute d squared, d action at a point like this of the function f. So what do we have to do? This is the second derivative of f composed alpha. So instead of writing f composed alpha, I already take the explicit expression of the function, and I substitute to the point p alpha of t. So now let's do it even without writing because this is what? This is alpha of t minus p0, scalar alpha t minus p0. First derivative, of course, is what? It's twice alpha prime scalar alpha of t minus p0. Remember, don't compute it at t equal to 0 because you want to do the second derivative. So this is d and d t at t equal to 0 of the first derivative at the generic t. So this is the scalar product alpha of t minus p0, scalar product alpha of t minus p0. So if I take the derivative of this, I get alpha prime scalar alpha t minus p0 plus alpha t minus p0 scalar alpha prime. So twice alpha prime of t alpha of t minus p0. So now we have to take the second derivative and evaluate at t equal to 0. So this becomes, if I take, so it's twice times what? So derivative of the first becomes alpha double prime of 0. So this twice goes outside times this evaluated at 0. Alpha of 0 is p by definition of alpha, so scalar p minus p0. Plus alpha prime at 0, actually here I should have added. So the h evaluated at a vector v. Plus alpha prime at 0, which is v, scalar alpha prime at 0, which is again v. So plus the norm square of v. This one, no, because when I take, so the question is, sorry, I have always to repeat, the question is, is this v or v minus p0? Well, no, because I'm taking the derivative. So p0 disappears because it's a constant vector. Now, the point is what is p minus p0? Now let's use the fact that it's a critical point, because this is true everywhere at any point. But now we have a relationship between p and p0 if we allow lambda to come into the game, and it's OK for us. So p minus p0 is equal to lambda. In fact, my minus p0 is equal to minus lambda n. So this becomes twice times the norm squared of v minus alpha double prime at 0 lambda n. In fact, lambda I could put lambda outside, and here I have n of p. But we know this. We have a geometric, I mean, an interpretation of a factor like this. As usual, it's Euler's theorem if you want. So the acceleration at that point of a curve, scalar product, the normal is exactly the second fundamental form. So this is equal to twice the norm squared of v minus lambda the second fundamental form at p at the vector v. Remember, here I use Euler's theorem for the geometric interpretation of the second fundamental form. So what can we deduce out of this computation? For example, corollary one, if s is compact, then there exists p in s such that p is elliptic, such that k of p, for a moment, k of p is greater or equal to 0. How is it possible? I mean, let's prove this, and then I will write a second corollary. Why is that? Well, the Gauss curvature is the determinant of the second fundamental form. So basically, you can translate that statement by saying that there exists a point where the second fundamental form is semi-positive definite. Now, do I have critical points on a surface for this function? I don't know, unless I put a topological condition. That's why I put compact. Because if it is compact, any continuous function has at least a minimum and a maximum. So there exists p1 minimum for f and p2 maximum for f. So at least I have two points where I can play this game. Otherwise, I cannot use this formula. Here I used, in the last step, I used the fact that I had a critical point. So at least I have these two special points. Now, which one will be good? If I have a minimum, how much is the action? Well, the action would be positive definite. Is it good for us or it's bad for us? I mean, if this object is a positive definite, in fact, strictly positive, but I mean quadratic form, is this OK? Well, it depends on lambda. Because out of this, of course, I take this factor. The two is irrelevant. I take this factor here, this addendum here, I put it there. And I get that lambda, the second fundamental form, is less than the scalar product. So as quadratic forms, I'm comparing the second fundamental form with the scalar product. OK? That's the way I want to prove that there exists a point where this is positive definite. So if I want to use this trick, I need to put this one on this side and then say that the second fundamental form is bigger than something, which is positive definite. The scalar product is clearly positive definite. OK? This is the idea. So where I can use this trick? How do I use this trick? Well, it's better to take the maximum. OK? If I take the maximum, but then you will realize that it's irrelevant. If I take the maximum, I have that it's negative definite d h, OK? So I can put the second fundamental form on the right. So I get something that looks nice, norm squared over vector less than lambda second fundamental form. So this implies what I want if I can argue that lambda is positive. So out of this, I get the norm squared of v is less than lambda, so at the minimum point. The norm squared of v is less than lambda second fundamental form. So now, why should lambda be positive? After all, in the geometric interpretation of a critical point, I'm saying something is a critical point if and only if p0 lies on the normal line. But who is telling me if it's above or below? Lambda means exactly this, the sign of lambda. But now think for a second. I can make lambda to be whatever I want, positive or negative. I mean, the value I'm not free to choose, but the sign I'm free to choose. Because if lambda is positive, I'm done. If lambda is negative, I reverse n. You see here, lambda positive, so n, the term is the sign of lambda. For one given choice of n, maybe I'm getting lambda. So if I choose n to be this one, then lambda positive means really it stays above. And in that case, I'm happy. But if for some reason p0 was here, I reverse n, and lambda becomes positive again. And the Gauss curvature does not change sign. That was an old observation we did. The principal curvature changes sign, change sign. The mean curvature changes sign, but the Gauss curvature doesn't. So now you understand that actually both were good. Because at the minimum point, by at worst switching the orientation, I can assume lambda is positive, and this inequality proves that the second fundamental form is positive definite. And I'm done. But at the maximum point, this was negative. I put it the other way, the other side, and I seem to get the wrong inequality, but I can make lambda to be the one I want. And I make it again the good inequality. So that's it. So at those two points, I get k greater or equal to zero. Okay, let's freeze this story here. And now, or in any case, but just remember, that this topological assumption was there only to guarantee the existence of a minimum of a maximum. If for some other reason you know that your surface, and there is a point for which your surface has a minimum and a maximum, those points will be of non-positive, of non-negative curvature. Very well. So now let me start with one of the biggest theorems we are going to prove in this course, which has a very weird statement. I mean, it's something that certainly doesn't look like a great theorem at the beginning. It seems a very technical statement. No, no, no, let's skip it. Otherwise, I cannot prove this theorem, which is long and big, okay? Corollary two will be an exercise. I mean, you can see how flexible this argument I gave you is. You can draw many conclusions out of this, okay? I just wanted to show you the line of the argument. So let me state and prove Hilbert's theorem. As I said, it's certainly, it's a statement which doesn't warm your heart at the beginning. You will have to wait half an hour or 45 minutes to realize it's really an important statement. Now, so you start with an oriented regular surface and of course you call, I mean, we keep on calling K1, K2 the principal curvatures. Oriented means it's orientable and I fix the choice of the normal vector n, okay? So oriented means it's a surface plus the choice of a normal vector. If I haven't done a choice of a normal vector, I have a choice, I mean, I know what K1 and K2 are, okay? Otherwise there is this sign problem, okay? But you fix n and at that point there is no ambiguity what are the principal curvatures. Then suppose there exists p, a special point on a surface, on the surface. Such that, one, K, the Gauss curvature at this point is positive, so suppose there exists a point of positive curvature. Two, suppose that K1 has a local minimum, so the small principal curvature has a local minimum at p. Three, K2, so the big principal curvature has a local maximum at p. So under these three assumptions p is umbilical. As I said, this doesn't look like a fantastic theorem, okay? We should be able to judge what is a great theorem by looking at the statement and this seems a very strange way to find umbilical. But remember umbilical means the two principal curvatures at p are the same, okay? Who cares? You will see, okay? Now, freeze the problem Y and let's try to prove it. Now we have our surface somewhere in space and of course all the things that we are doing in the geometry of surfaces is invariant by rigid motions of R3. I mean if I take a surface and I translate it in space, of course principal curvatures mean curvature and Gauss curve, everything will be the same at the corresponding points. If I take a surface in space and I rotate it, okay? Principal curvature, mean curvature, Gauss curvature will be the same, okay? Everything, so our geometry is kind of or O3 plus translation invariant in R3, okay? So just for convenience, we can assume that our special point, you know, we start with the surface with the special point actually is actually the origin of R3. We can suppose p is the origin and this would be just a translation in space. And then thanks with the rotation, I can also assume that the normal vector at this point is for example the standard vector along the z-axis, okay? So it's 001 and this would be kind of a rotation only of one axis. I really want to make full use of the rotations of R3. I can do something more. I can say that the principal directions at this point, so this is something which line the tangent space, okay? With the rotation of R2 keeping this normal line fixed, I can put them in the position I want. This is an orthonormal basis of a two plane, okay? So with the rotation of this plane, I can put it for example in the standard position. So I can assume E1, the standard vector 1100 and E21010 are actually the principal directions. Now unfortunately I really need the whole blackboard. So basically what it means we have put our surface in the position where the original point has become the origin. The tangent plane at that point to the surface is the xy plane. The normal vector then, of course, it's forced to be this one or minus this one. I suppose it's this one going up, okay? And the two standard vectors of R2 are the principal directions, okay? This is achieved with translations and rotations of R3. So nothing has changed in the geometry, in the curvatures and everything, okay? Very well. Now, we have our surface somewhere here, okay? So now this is more or less the picture. But now either as a corollary of the implicit function theorem, or if you want you can use this and the previous discussion we made about the height function, we know that locally there is at least a neighbor of this point here of the point P which is the origin where the surface is a graph over the tangent plane. Of course the tangent plane in this case is the standard R2, okay? So around P, S is locally the graph of a differentiable function, F, okay? So that means that there exists a function defined around the origin of R2 with values in R such that S becomes the graph U, V, or if you want X, Y, F of X, Y, okay? So now let me try to understand which are the properties of this F. Well, let me translate everything we said here in terms of F, okay? So all these properties translate into what? F of zero, zero is zero, okay? Because I want the origin to go to P, okay? So P is zero, zero, zero, so F must be zero at this point. Then what else? We can also assume that the partial derivative, what are the standard basis of the tangent vector? So if you want, you can construct now a local parameterization around this point of the form graph, okay? So the standard basis with respect to this parameterization of the graph, of the tangent plane to the graph is the vector X, U, X, V. Okay, so it's one, zero, F, U, and zero, one, F, V. I can certainly play the game in such a way that F, U at the point zero, zero is zero because now I'm writing the fact that the tangent plane is the horizontal plane, okay? At that point, only at the origin, the tangent space is the horizontal plane. So that means that the components of X, U, and X, V in the vertical directions are zero, okay? So F, U at zero, zero is zero, and F, V of zero, zero is zero, okay? So now what I've written, I've written this property and all together the fact that the tangent space is the horizontal and then of course the normal is this one, they come together, okay? Very well. So basically that means something more geometric also. It means that the direction U, so, and of course since this is true, what is X, U at zero, zero? Well, writing in this language, it becomes E, one, okay? So X, U is automatically not just any tangent vector, but it's the principle, the first principle direction. And X, V is the second, okay? This comes for granted. Now let me write all these assumptions in terms of F. Okay, now I have three hypotheses. So that means compute first and of course how do I express K and K1 and K2? The only way I can do it, if I have a local chart, I express it, I have to compute all the coefficients of the first and the second fundamental form, okay? But this is a graph, so it's very easy, okay? In fact, okay, let's do it. But I mean, if, what? No, otherwise there is no time, okay? It's actually simpler than the examples we did. So E, little E, if you make the computation becomes F U U divided, probably actually we did something like this, plus F U squared plus F U P squared, okay? F, little F, oh sorry, little F now seems fighting with the previous notation. There is no confusion, okay? One thing is the function, one thing is the coefficient of the second fundamental form. So little F becomes F U V divided by the square root over the same thing. I mean, this object that I'm writing below is always E G minus F squared capital, okay? And that's what happens above. And little G is F V V divided by square root plus F U squared plus F V squared. And these are true on the whole neighborhood. So not just at the point zero, zero, okay? These are general formula for the coefficients of the second fundamental form of a graph. Everywhere, everywhere covered by the graph chart, okay? Now, what do I know about these functions? Well, the other thing I know, since U is essentially corresponding to the first principal direction and V is corresponding to the second principal direction, the other thing I know is that F U U at the point zero, zero, nowhere else, I mean, I have only an assumption in zero. So at the point zero, zero, this is nothing but K one of P, okay? F of U V at zero, zero, you see? The point I'm saying is that the coordinates U V really correspond to the principal directions. But that means that the second fundamental form is in diagonal form at that point, okay? And of course, the first entry becomes the first eigenvalue. The one outside the diagonal becomes zero. And the other on the second entry, second, second, second line, second column is just K two, okay? So this is again coming from having required that these are the principal directions, okay? Okay, now let me define two curves which are kind of instrumental. They don't carry any particular meaning except they are useful for doing the computation. So define two special curves on the surface like this. Alpha, and I call it the parameter U, to be just the image of the first coordinate function. So X of U zero. Basically, if you imagine this to be the graph, actually what I drew is globally a graph, but for what we know in general, this picture is true only locally. So basically if this is the coordinate U and this is the coordinate V, I'm taking this line because I'm taking V equal to zero, this line here, and I'm looking at its image on the surface. This is alpha, okay? And then of course I do the other one. I take U equal to zero and V three. So that means I take this line and I go and see which is the image on the surface and I call it beta. And then I define another two curves. E and I call capital E one. So in fact these are in some sense are vector fields. I mean E one as a function of V, so this is kind of defined only for a parameter V to be one over the norm of X U at zero V times X U, so it's the unit vector in this direction. And symmetric, so you see now I'm playing kind of a dual game because I'm taking it as a parameter V, but I take derivatives with respect to U and I put U equal to zero. So now I do the opposite game. E two of U and then of course this will be one over X V, the norm of X V at U zero, yeah, at U zero times X V at U zero. These are just definitions, okay? You will see why they are useful, but I cannot really convince you. But which are the properties of all these things? So for example, E one at every point, E one is what, is a tangent vector to the surface at the point of the form beta of V, okay? And duly this object here is a vector but it's not any vector, it's a tangent vector to a point alpha of U. That's just because the tangent space at every point is spanned by X U and X V, no? So the only place that since you are computing at this point, this is exactly the point beta of V, no? And this is a tangent vector but it's computed at U zero which is the image of, I mean, alpha of U, okay? And they are just normalized, they are of norm one. So, okay, unfortunately I have to erase the hypothesis but the computation is still, yes? No, so the question is you find it quite uncomfortable set that the small one as a minimum and the big one as a maximum. It seems more intuitive to ask kind of the opposite. Now if you want the two, no, it's a right observation. It's a right observation, I mean, you have two functions. One is above the other and somehow the conclusion is that at that point they are the same. So basically you seem to, you would like to push the small one up and the big one down, no? And he's asking exactly the opposite in some sense. If you push the small one down and the big one up, they are the same. Okay? So I think you got why it's an interesting theory because it has a non-intuitive hypothesis, okay? Now, let's keep that one and let's go on here. Having defined all these things, we can define two special functions. Let me call it so. I keep on defining objects. I call it H one of V. To be what? To be the second fundamental form at the point beta. So since I have a special tangent vector, special, I don't know why exactly. You will see in a minute. I have a special tangent vector at the point beta of V. Let's compute this second fundamental form, okay? So of the vector E one, E one twice, okay? And let's define H two to be kind of the, again, the usual as a function of U, the second fundamental form at the point alpha of U of E two, okay? And here I write it only once because I don't have space. So you count it twice as usual in our notation. But we have an expression for the coefficients of the second fundamental form, which is good for all points in this neighbor, not just at the origin. So we know how to compute those functions. For example, H one of V in terms of the coefficient because you see E one is a multiple of X U, okay? So the second fundamental form evaluated at X U, X U by definition is little E, okay? So how much is the function H one? Well, it's exactly F, that expression, F of U U, F of U U divided by the square root of one plus F U squared plus F V squared. Now I have just to be careful that E one was not exactly X U, but it was normalized. Of course, this is a quadratic form. So if you multiply a vector by something, the something goes out, but it is multiplied twice. So it goes out to the square, okay? So it's divided by the norm squared of this function, of this vector X U, okay? But this is another thing I can express easily because X U, well, we wrote it only at zero, zero, but of course in general would be the vector one, zero, F U, okay? So the norm, the norm squared of this vector is one plus F U squared, okay? All these evaluated at the point, not at every point, but only at the point zero V by definition. And same way, H two of U, now this is the second vector in my order basis of the tangent space. So this becomes the second coefficient, okay? Of the second fundamental form. So, and for the same reason, this is F V V divided by the square root of one plus F U squared plus F V squared times, now I have to divide by the norm of X V squared, okay? X V is zero one F V, so it's norm squared is one plus F V squared, okay? And this is computed only at the point U zero, okay? Why we have done all this mess? It's clearly complicated objects, but now you see why it was convenient. How much is H two at the point zero? Well, by definition, it's just the second fundamental form at alpha of zero, alpha of zero is P, because if you put U equal to zero, you are getting X of zero, zero, so that's P. So this is the second fundamental form, is equal to the second fundamental form at P of what? What is the vector E two of zero? Well, E two of zero, it's small E two, okay? Twice, and with our notation, I don't duplicate, okay? But what is the second fundamental form? How much is the second fundamental form at P of in the direction E two? E two was the principal direction in the corresponding to the eigenvalue K two. So this is actually K two of P, but K two has a maximum. So now let's start using hypothesis. Everything we set up to now was completely general, no? Now, K two has a maximum at P. It's the big one and still we are asking it as a maximum, contrary to, so that means that this is greater than or equal to K two at any point alpha of U for U sufficiently small. But this is also greater than or equal. So now to the second fundamental form at the point alpha of U of to the vector E two of U. Because this is the maximum of the eigenvalues. So if I compute the second fundamental form at the same point twice on any vector, I have to get less or equal. They are equal if and only if this vector stays the principal direction. I never said that. I know it only for U equal to zero. For U different from zero, I don't know if this is the principal direction. So I'm not saying that this is equal, but certainly K two will be bigger, okay? Because it's the maximum of the values of the second fundamental form. On vectors of norm one, remember, this I'm using the standard interpretation of the eigenvalues of the quadratic form has the maximum or minimum, on the circle of vectors of norm one. That's why I have divided by the norm because that's the only thing I can claim, not X U, okay? So here things are coming all together. And you realize that there was no way to simplify this object. They look bad, but I mean that's the best you can do, okay? Okay, but what is second fundamental form at alpha of U or at E U is by definition H two of U. Now this is really equal because this is the definition. So what have we observed? We have observed that H naught, H two of zero is greater than or equal to H two of U for any U. So in particular, zero is a maximum, local maximum for the function H two. So zero is a local maximum for H two. Let's see if also H one has some strange property like this, which actually justifies all the definition. How much is H one of zero? Well, H one of zero, what it is, here it is. H one of zero is the second fundamental form. Again, at beta of zero, but beta of zero is again P, okay? Is the second fundamental form at P of the vector E one of zero, but E one of zero, that's where E one is here. It's X U at zero, zero. But X U at zero, zero is E one. So we start in the same way, except we change E one and D two. But how much is this? Again, this is K one. And what are we asking on K one? It has a minimum. So this is less than or equal to K one at any point of the form beta of V. K one is the minimum of the quadratic form. So this is less than or equal to the second fundamental form at the point beta of V of E one of V. Again, by the same interpretation of the eigenvalues. But this is by definition H one of V. So what have we learned? So zero is also a local minimum for H one, okay? So the summary of all this is that we found these two functions and we have the property, I can express this property by second derivatives. So in particular, I know that so H two, the second derivative of H two, even though now primes will be with respect to different parameters because H two is a function of U and the other is a function of V, but it's always one parameter. So H two at zero, this is less than or equal to zero because it's a local maximum. And H one double prime is greater than or equal to zero. Okay, very well. Now there is still a painful step because now I really want to see if this inequality is really interesting or not. Up to now it seems like a game that we are playing. Let's compute H two prime of U for example. So we have also, so you see here, we didn't use the explicit expression for F. I want to write down these inequalities in terms of F, okay? Unfortunately, what it means is that I have to take the second derivative of this with respect to U and express it in terms of all this mess. Be patient, okay? If I want, at the end, I would like to make the second derivative. Let's start with the first. H two prime and then, but I don't have to evaluate it as zero because I'm going to make the second derivative. Well, how much is this? Well, let me group everything in this way, okay? Because of course I use the product rule here. So this becomes minus one plus Fv squared to the power minus two times twice Fv because if I start taking the derivative, so that means in my notes, I started taking the derivative first of this. So I take the derivative of this times everything else. So the derivative of this with respect to U is what? Is this subject to the minus, is minus, this subject to the minus two times twice Fv, FvU, okay? So this is the derivative of one over this. But then I have to multiply by everything else. So, but let me write instead of denominators, now I prefer powers. So FU squared plus Fv squared to the power minus one half, okay? Times FvV, which is still there, okay? FvV, so that's the derivative of this times this. Now, let me take the derivative of this times everything else, okay? So then I still have a minus. It's better to, they are long formula. So in my convention, changing minus minus not a plus, or minus means minus, sometimes, okay? So that means I'm not touching these two factors and I take the derivative of this with respect to U. Well, okay, that means, so that's something to the minus one half, okay? So that becomes, it becomes minus one half, that object, one plus FU squared plus Fv squared to the power minus three halves times, well, okay, let's freeze it. We still have to make the derivative with respect to what's inside, but this is one plus Fv squared to the power minus one, FvV, I'm not touching and I still have to take the derivative of what's inside the square root with respect to U, okay? And what is the derivative of what's inside U with respect to U? It's twice, twice FU, FUU plus twice Fv, FUV, okay? So it's twice, FU, FUU plus twice FV, FUV. And this is the second thing, so we have taken the derivative of this, derivative of this and then I kept the easiest last, so I can relax derivative of this times everything else. So plus now, plus. But let me write everything in powers again, FU squared plus Fv squared to the power minus one half, one plus Fv squared to the power minus one and then just the derivative of FVV, so it's FV, VU and that's it, okay? At the point U, zero. Now, you see how beautiful this is. Now we have to take the second derivative. So this is the moment where you change subject, you decide you want to study biology or you decide that there must be a trick, okay? So instead of doing it with your brain, oh, you can do it, of course. But I mean, unfortunately I just erased, but I mean I needed space and I erased a bit too early. We have a lot of vanishing at the point zero. Now, this is the formula at the point U, but then at the end we are going to put U equal to zero. And at U equal to zero, we have, in fact, maybe it's better to write them somewhere. What do we know? Remember, we knew FUU, let me write it here, FUU00 is equal to zero, FUV00 is equal to zero, FVV of 00 is equal to zero. And about the first derivatives, again, FU at 00 is equal to zero, FV at 00 is equal to zero, and F at 00 is equal to zero. So once you remember that we know all this, oh, sorry, sorry, sorry, sorry, sorry, sorry. K1 and K2, okay? K1 at P and K2 at P, okay? But you see, once you remember this, you don't change subject yet, okay? Okay, because, of course, imagine the procedure to compute the derivative of this, okay? But without even writing, let's start. Is there anything, how can anything survive after the derivative when I put U equal to zero? Because you see, if I, so this, let's take the derivative of the first line, okay? Well, of course I should do. Derivative of this times this. Well, F, but there is an FUV, a beautiful FUV, which has zero will be zero, so I don't do it. So derivative of this, if I take, same thing, if I take the derivative of this times everything else, there is still this beautiful FUV, okay? Very well. So in principle you might say, well, so of course the derivative of this, there is written nowhere. I have no information about third derivatives. So in principle there could be only. Third derivative of this times everything else. And actually this seems to stay, no, because there is an FV, okay? So you see that even without doing any computation, and then for this and this, there is still either one of them. So I'm fine, the first line is zero. Derivative of the first line would be zero, and I don't have to do it. Okay, so let me group here, this is H2, double prime at zero, okay? Let's see what survives, because actually very few things will survive. But hopefully something will survive. Okay, let's see. Let's play the same game on the second line. Well, on the second line, F here, this, because you see at zero, this is one. So it's not very useful. This is one, it's not useful again. This is K2, which I don't know if it's zero or not. I have no information, so it could be zero. And how much is this? Well, this is zero for two reasons, okay? Both the first and the secondary, sorry, because there are the first derivatives, okay? That means that if I keep on taking derivatives of this, this, this, I'm wasting time. The only thing which is interesting to do is the derivative of this times everything else. And not just that, if I take the derivative, when I take the derivative of this, the only thing which is clever to do is to take directly the derivative of this and this, but not of the second derivatives, okay? And when I evaluated u equal to zero, what does it happen? Well, so this becomes minus one-half. Let's see what survives. This evaluated at zero, so one. This evaluated at zero. So one, this becomes FVV at zero, so K2 of P. And now, there is a two here, okay? And now, I know that I have to take the derivative of this with respect to u. So this becomes FUU times FUU, okay? So in this language, well, actually, it would have been better to skip it. Sorry, before substituting K1 and K2, leave them as derivatives of F. So actually, that one was FVV, okay? So FVV is zero times derivative of this. So FUU times FUU, both at zero, okay? Plus, plus nothing, because this is an FUV, okay? So even if I take the derivative of this, this will kill them, and vice versa, okay? So that's the only thing which is surviving in the second line, okay? Now let's go to the third line. Seems amazing that I'm not doing mistakes, but, okay? Second, the third line, same principle. Well, now it looks a bit more complicated, because you see I have no informations about this. This is one, and this is one in the origin. So what do I do? Well, now I have to imagine that I play, so in principle, I should do everything. But without even writing, how much would be the derivative of this? Well, this would be minus one-half, this subject to the minus three-halves, which is evaluated at zero, which is one, so in principle, it's there, so it's actually minus one-half, okay? Times what? Times the derivative of FU squared. And this squared is fantastic, because this becomes two FU, FUU. So there is an FU, so it's zero. And when I do it here, it's of course two, FV, FUV, so there is FV, so it's zero. So now I play the same game, but I have to do one step more. So the derivative of this will be zero anyway, okay? So I don't do it. Let's see what happens here. But again, this will be minus one, this subject to the minus two, twice FV, FUV, so there is FV. So the key point is of course that these are quadratic, okay? So when you take the derivative, there is always a linear thing in front. The linear part is zero, so, okay? So also the derivative of this doesn't matter. Throw it away. So we are left with this, okay? And this of course should be there, because this is one, this is one, and this is, so, plus, with the sine plus, so here no bracket, minus this object, plus one, one, FVV, UU, at zero, okay? So it was not that terrible. Now we play, remember, remember the logic of the proof. We got this inequality here, and we want to express it in terms of F. So we are halfway through, because now we know what is this with respect to F. Four derivatives in F, but it's something, okay? Now we should do exactly the same thing for this. Clearly, I'm old enough not to do it, but you are young enough to do it. I prefer not to comment on symmetry properties of these functions, okay? So I can tell you this becomes minus FVV, FVV, FUU, at zero, plus the same thing, because actually they would arise as FUU, FVV, but then of course, Svart's lemma, the order of the partial derivatives is irrelevant. So it's FVV, UU, at zero, okay? So it's true. I mean, what you were suggesting, it's true, okay? It's symmetric in everything. What do I get out of this? Well, in which order do I want to FVV, FVV? So I look at H2, H2 double prime at zero minus H1, because of course the nice thing is that there is a part in common. So I take a difference, but it's better to take this one. What do I know about this? First I know that this is negative, well, non-positive, okay? But on the other hand, I can express it now as what? It becomes this minus this. So there is an FUU, FVV in front, evaluated at zero times with the plus, I have to take this. So FVV at zero minus FUU at zero, okay? And now you still don't know why, but you are very happy because now you substitute the geometric interpretation of these objects as we were doing at the beginning. I suggested to do it at the end. What is this? This is the Gauss, this is K1 times K2. So it's K times K2 minus K1. And what have we learned here? Is that this is non-positive. So this seems the typical theorem. Well, you do a computation because it seems an interesting computation. You don't know why, you get to the end and then you put the NDA hypothesis in the theorem to make it work. Because now, how is it possible that the product of two numbers is non-positive? Well, if the first one is positive, the second one must be negative, but how is it possible? This is big minus small. So the only possibilities actually, they are the same and this is zero. So this implies the theorem. I leave you a few seconds to finish copying and thinking about this proof because now I have to convince you it was worth spending 40 minutes of your life on this theorem, okay? Because now, corollaries. Corollary. Which actually was the original reason why Hilbert, this is Jeunet-Liebman. It was the original proof of this theorem was very complicated and Hilbert came up with the, well, you can see, I think it's fair to say it's a simple proof because it's quite elementary. It's just, it's long and you have to make many derivatives, but I mean, you are not really doing anything, particularly sophisticated. So that you can consider the previous theorem kind of an elementary theorem. It's just, it's a long theorem, okay? So what does this theorem says? Well, a compact, suppose S is compact and connected. K is positive everywhere. Moreover, assume, suppose H is constant. Then, suggestions. What could it be? That's a good suggestion. Well, the sphere certainly satisfies everything, no? In fact, it satisfies even more because this is not just a surface with positive curvature, but it's actually constant positive. So it's even, no? And, well, of course I cannot expect another proposal because the theorem says exactly this. Suppose H is constant, then S is a sphere. And in this case, it's not, I don't have to be worried about, it's a piece of the sphere, because it's compact, okay? So it has to be the whole sphere. Now, how to prove it? All right, of course sooner or later, Hilbert's theorem will come in, but first there is a little delicacy. First claim, well, let's get to the claim. So let's see, be a real number such that, now here I would like you to think that this is a strange statement for a moment. But in some sense, it makes sense. So I'm assuming that the mean curvature is constant, okay? And now I'm giving a name to the value of this constant. Okay, that's okay, but, well, first, let me argue that C, then the first little claim is that C is non-zero. I don't know if it's positive or negative, nobody's telling me anything about the sign, but certainly it's non-zero. Why? Well, because if it was zero, you contradict the Gauss curvature assumption. Because otherwise, of course, this can happen only if you have a millical point. K1 is equal to K2, okay? So K1 is equal to minus K2, sorry, not umbilical, but, okay? Otherwise, K1 is equal to minus K2 because the sum has to be zero, okay? But then the product cannot be positive, okay? And we have exactly the opposite assumption. But now, the slight delicate claim is that if you have a surface whose mean curvature, in fact, is never zero, but I mean, suppose it's a constant, non-zero, then S is orientable. Meaning I can choose a unit normal vector field defined over the whole surface, that means orientable. Remember, locally I can always do it, but globally there are problems, okay? Orientable means no, there are no problems even globally, but now this is a bit delicate. There is not much to write. In fact, probably whatever I write makes more confusion than you have to think of what it means. Now, of course, around any point, I can define N. I have a general theorem that tells me if I have a chart, on this chart I have N. So the problem is not local, it's automatically global. Okay? So, suppose at a given point, I choose N in such a way that H, with this choice of N, is equal to C. So now this is where you have to think that I'm not cheating. Because I said H is equal to C, but H depends on the choice of N, because otherwise it becomes minus C. So if you give me one choice, it's C. If you give me the opposite choice, the only other possibility is minus N. I get minus C. So when I decide that H is equal to C, I can do it, but then in some sense this forces the normal vector to be one of the two possibilities at every point. There is only one of the two for which H is equal to C. You agree? Fortunately, because C is non-zero. This is where you use, because of course, zero is the only number which is equal to minus itself. If C was zero, my argument would collapse. Okay? But see, once we have observed that C is non-zero, this is the term is a choice of N. But this N has to be globally defined. Because if it stops to be globally defined at some point, it means at some point it has become minus N, but then H would have become minus C. It looks like a trick, I know. You have to think that the logic is correct. Okay? So there is really nothing to write, except remembering that the choice of the value of the mean curvature, the term is a unique normal vector for which that is the value. Okay? Once you do that, that's okay. Well, of course, no, it's okay. Why this is important? Because I want to speak, so okay, so this is done by this implicit argument. If I have an orientable thing, I have a choice of N everywhere. The Gauss curvature, the principal curvature, the mean curvature, everywhere become globally defined. So this is the advantage of the important thing. So I can think of K1 and K2 become really function over the whole S. This is the way I want, I use it. Otherwise, around every point, I have to take a chart, a normal vector, and compute them. But if I change charts, they change. On a non-orientable surface, I cannot speak of K1 globally. Do you agree? So this is kind of the point, okay? Because we also know that these functions maybe are not really differentiable everywhere. But certainly are well-defined now, thanks to this, and are continuous. Are well-defined and continuous. And that's enough to use topology, okay? Because remember, now try to put Hilbert's theorem into the game, I'm looking for a point where K1 as a minimum and K2 as a maximum and K is positive. Actually, K positive, I've put it everywhere. So now I take P such that K1 has a local minimum at P. This exists because now I'm using this. So you see now, one by one, I've already used this. This actually I used it implicitly in this argument of orientability. Now, compactness tells me that this function has a minimum. Now, this point, is this good enough to use a minimum at P? Now, this point, is this good enough to use Hilbert's theorem? Well, this satisfies one property by brute force. The other property by hypothesis, there is one missing. Is K2, has K2 a maximum at this point? Yes, and now I use again this, okay? Okay, now P is automatically a local maximum for K2 because H is constant. So the sum of K1 and K2 is constant. One as a minimum, the other must have a maximum to balance the other one because H is equal to C. But then what does Hilbert's theorem tell me? Well, not much actually it seems because it seems that this point is umbilical. But actually if I want to prove that something is a sphere, the only thing I have, the only gun I have is to prove that every point. And this seems to be a very special point. I mean, come on. This is certainly not the generic point it seems. But no, actually take another, take Q in S, any point. Now, how much is K2 of Q? Well, P, well in fact it's a global maximum. I mean, also this one, okay? To use Hilbert's theorem it's enough local but here I'm using already automatically global maximum, okay, global minimum, global maximum because I'm using compactness, okay? So, wherever Q is, K2 of Q is less than or equal to K2 of P. But this is equal, now P is umbilical. So this is equal to K1 of P but P is a minimum for K1. So this is less than or equal to K1 of Q but this is impossible unless they are all equalities and Q was any point. Okay, so our surface is covered by umbilical points. Every point is umbilical, that's it. Now, in the last 10 minutes, now you can see this proof is actually very flexible. And the most famous actually, this was the origin of Hilbert's theorem. So, but actually Hilbert himself proved another corollary of his argument which in some sense it's even more interesting and this has passed to history, has Hilbert Liedmann theorem which says that if a surface, if S is compact and connected, if the Gauss curvature is a positive constant, remember before we did not require it to be a constant and it turned out to be a constant. We require it to be positive but not in principle a constant. Now we require it to be a positive constant. Then, again, S is a sphere. So spheres are actually the, because now of course what does a geometer in this sense when studies this subject that? You have defined many functions, K1, K2, H, K. You would like to know, you would like to classify surfaces by looking at these curvatures. So what does it, of course these are functions so the simplest question you can ask is which are the surfaces for which one or two or three, one of these functions is constant? Okay, it's the first question in a general classification theorem, okay. So if this constant that you have chosen is a positive number, Hilbert tells you it's a sphere, okay, nothing else. Then of course we will spend a couple of lectures to decide why this is the most important curvature, okay. But that's another story. Now the argument is very similar to the one below, to the one of the previous corollary, okay. First, oh sorry, actually, okay this is true. What I said is true but actually you don't need this. This is important. In particular you cannot produce flat compact surfaces. Okay, I'll come back to this theorem commenting but once we understood the meaning of K in another way. So in fact the first thing to observe is that K must be positive. So this constant, the first observation is that if you have a compact surface with constant curvature, Gauss curvature, it has to be positive. Well we observed at the beginning of the lecture that every compact surface has a point whose Gauss curvature is greater than or equal to zero. So this, by that observation, corollary one before, by that observation this constant cannot be negative. Now you have to improve it a little bit that argument to show that this constant is actually positive, exercise because it's a minor improvement of what I said in the proof of the corollary, okay. So you cannot get zero. So this comes from the beginning of the lecture. And now, now you see I would like to play the same game. So I would like to look at the functions K one and K two globally on S and then to start arguing as before. So first I need to argue that the surface is orientable. Otherwise these functions are not global functions and then if these are not global functions, I cannot use compactness to tell you pick a minimum and pick a maximum, okay. Which is on the contrary the only way in general I can control critical points, okay. So why this should be orientable this time? Well I want to play the same game as before. If I can prove that the mean curvature is non-zero everywhere. You remember in the previous argument it was not important, it was a constant. The important thing is that it's non-zero, okay because you decide which, so positive or negative and then the problem is crossing from positive to negative because N and minus N differs by minus one, okay. So you don't really care that C is a constant in the argument before, okay. So why this and I claim that this comes from this. So H is non-zero everywhere. Well but it's the same argument as before. If it's zero K one is equal to minus K two and then the Gauss curvature is minus K one squared, okay. So it's less than or equal to zero but this is strictly positive. That's it, game over, okay. Then the same argument as before S is orientable. Then I pick one N, one of the two possible N and then I have global functions K one, K two defined on S, okay. And then again I pick a minimum P minimum for K one for example, so it satisfies one of the hypothesis. The other one is for free because K at P is positive because it's positive everywhere. In this case it's even constant but we don't care at the moment. And now I have to argue that this, the same P is a maximum for K two. But again, it's algebra, okay. Which said by a geometer means it's something that we are not even supposed to think about, okay. We know that K one times K two is a constant function. So again, if I pick a minimum for this automatically I must have picked a minimum for that, okay. So this automatically means that P is a maximum for K two. But then we are again in the same situation as before. P satisfies all the hypothesis of Hilbert. So it's umbilical but again only P is umbilical. And then you play the same game, okay. The same thing I wrote before tells you that every point is umbilical, okay. So in fact this was, in fact you will find it in books probably as Hilbert's theorem on this one. But so classification of surfaces, of compact surfaces of constant Gauss curvature. I hope in four lectures I will tell you why this is not just a nice theorem. I mean this is really a great theorem, okay. For reasons you will understand in a couple of weeks. So I'll come back to this, okay.